A) \[1.0\,\Omega \]
B) \[0.5\,\Omega \]
C) \[2.0\,\Omega \]
D) zero
Correct Answer: A
Solution :
Key Idea: This problem is based on the application of potentiometer in which we find the internal resistance of a cell. In potentiometer experiment in which we find infernal resistance of a cell, let E be the emf of the cell and V the terminal potential difference, then\[\frac{E}{V}=\frac{{{l}_{1}}}{{{l}_{2}}}\]where \[{{l}_{1}}\]and \[{{l}_{2}}\]are lengths of potentiometer wire with and without short circuited through a resistance. Since, \[\frac{E}{V}=\frac{R+r}{R}\][\[\because E=I(R+r)\] and V = IR] \[\therefore \]\[\frac{R+r}{R}=\frac{{{l}_{1}}}{{{l}_{2}}}\]or\[1+\frac{r}{R}=\frac{110}{100}\]or\[\frac{r}{R}=\frac{10}{100}\] or\[r=\frac{1}{10}\times 10=1.\Omega \]
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